214 XIII INTEGRATION belongs to some Un, which intersects only finitely many of the Vfc), and we have |/*|*(/) =+oo, because M*(/) £ £ |/i|(/i) for all /i. fc= i If S = Supp(/*) is compact, the first part of the proof shows that \fj\(f) £ M(S) ' sup |/(jc)| for all functions/e #C(X), and therefore ju is a xeS continuous function on the Frechet space #C(X). Conversely, let A be a con- tinuous linear form on this space, so that there exists a compact set K a X and a constant c> 0 such that |A(/)| ^ c • sup |/(jc)| for all/s «C(X) (12.14.6). If xeK L is any compact subset of X, we then have |A(/)| ^ c • ||/|| for all /e Jfc(X; L), and hence the restriction u of A to Jf C(X) is a measure on X. Moreover, if A is a continuous mapping of X into [0,1], with compact support and such that h(x) = 1 for all xeK ((3.18.2) and (4.5.2)), then MJh) = A(/) for all functions/e V€fX)9 because/-//* is zero on K. From this it follows immediately that the support of \JL is contained in Supp(/z); hence every/e *C(X) is /i-integrable, and n(f) = j*(fK) = l(fh) = MJ). Q.E.D. (13.19.4) If 7i: X->X' is a homeomorphism and fi is a measure on X, we have Supp(7i(/0) = 7r(Supp(//)). This follows directly from the definitions. PROBLEM Let /x be a positive measure on X, and A a /n-measurable set. Let i(A) be the set of points xeX such that there exists a compact neighborhood V of x in X for which py n (X - A)) = 0. Show that /(A) is open and that /u(/(A) n (X - A)) = 0. (Con- x-A • ft).) 20. BOUNDED MEASURES For every (complex) measure /* on X, we define (13.20.1) Ml- sup which is a finite real number, or + oo. (1 3.20.2) We have ||/i|| = |/*|*(1). Hence \\IJL\\ is finite if and only if the positive measure |/t| is bounded (13.9), and in that case total mass ofX. with respect to /£.ce (Vn) of relatively compact open sets, as follows: